Accounting Solver

Solve accounting problems step-by-step with clear explanations. Upload a photo of your accounting question or type it in, and get instant, accurate solutions. Ideal for students and professionals needing help with accounting concepts, calculations, and journal entries.

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Use Cases

Discover how this template can be used in various scenarios

Accounting Students

Solving textbook problems, assignments, and exam questions with step-by-step accounting explanations.

Professionals

Quickly checking accounting calculations, journal entries, and financial statement impacts.

Small Business Owners

Understanding accounting concepts and preparing basic entries for business transactions.

Tutors and Educators

Demonstrating accounting solutions and concepts in class or one-on-one sessions.

Job Seekers

Practicing accounting problems to prepare for interviews or certification exams.

How to use the Accounting Solver

Accounting Solver Templates & Examples

  • INPUT

    You are considering investing $10,000 at an interest rate of 4% compounded annually for five years or investing the $10,000 at 5% per year simple interest for five years. Which option is better?

    OUTPUT

    Analysis
    This problem compares two ways to grow a cash investment over five years: one using annual compound interest at four percent and the other using simple interest at five percent. The approach is to compute the future value under each method and then compare which is larger.

    Solution

    Step One – Compute future value under compound interest
    Operation uses the compound interest formula
    A=P×(1+r)nA = P \times (1 + r)^n
    where P is the principal, r is the annual rate, n is the number of years.

    Calculation of growth factor
    First compute the square of 1.04
    1.042=1.08161.04^2 = 1.0816
    Then compute the cube of 1.04
    1.043=1.0816×1.04=1.1248641.04^3 = 1.0816 \times 1.04 = 1.124864
    Then compute the fourth power of 1.04
    1.044=1.124864×1.04=1.169858561.04^4 = 1.124864 \times 1.04 = 1.16985856
    Then compute the fifth power of 1.04
    1.045=1.16985856×1.04=1.21665290241.04^5 = 1.16985856 \times 1.04 = 1.2166529024

    Multiply by the principal
    A=10000×1.2166529024=12166.529024A = 10000 \times 1.2166529024 = 12166.529024

    Rounded to cents
    Acompound12166.53A_{\text{compound}} \approx 12166.53

    Step Two – Compute future value under simple interest
    Operation uses the simple interest formula
    A=P×(1+r×n)A = P \times (1 + r \times n)

    Substitute values
    A=10000×(1+0.05×5)=10000×1.25=12500A = 10000 \times (1 + 0.05 \times 5) = 10000 \times 1.25 = 12500

    Thus
    Asimple=12500A_{\text{simple}} = 12500

    Step Three – Compare results
    We have compound interest outcome
    12166.5312166.53
    and simple interest outcome
    1250012500
    Since 12500 exceeds 12166.53, the simple interest option yields the higher future value.

    Answer
    The five-year simple interest investment at five percent yields 1250012500 which is greater than the compound interest result of 12166.5312166.53. Therefore the simple interest option is the better choice.

  • INPUT

    A project is expected to generate a cash flow of $2,000 in year 1, $800 in year 2, and $1,000 in year 3. At an interest rate of 10%, what is the maximum amount that you could invest in the project at year 0?

    OUTPUT

    [Analysis]

    This is a capital‐budgeting problem requiring the time‐value‐of‐money method. We must find the present value of each future cash flow at the 10 percent discount rate and sum them. That sum is the maximum amount to invest at year 0.

    [Solution]

    Step one
    Define the present‐value formula:
    PV=CF1(1+i)1+CF2(1+i)2+CF3(1+i)3PV = \frac{CF_{1}}{(1 + i)^{1}} + \frac{CF_{2}}{(1 + i)^{2}} + \frac{CF_{3}}{(1 + i)^{3}}

    Step two
    Substitute the cash flows and the interest rate (i = 10 percent):
    PV=2000(1.10)1+800(1.10)2+1000(1.10)3PV = \frac{2000}{(1.10)^{1}} + \frac{800}{(1.10)^{2}} + \frac{1000}{(1.10)^{3}}

    Step three
    Compute each discounted cash flow: • Year one term 20001.10=1818.18\frac{2000}{1.10} = 1818.18 • Year two term 800(1.10)2=8001.21=661.16\frac{800}{(1.10)^{2}} = \frac{800}{1.21} = 661.16 • Year three term 1000(1.10)3=10001.331=751.31\frac{1000}{(1.10)^{3}} = \frac{1000}{1.331} = 751.31

    Step four
    Sum the discounted amounts:
    PV=1818.18+661.16+751.31=3230.65PV = 1818.18 + 661.16 + 751.31 = 3230.65

    [Answer]

    The maximum amount that can be invested at year 0 is 3230.653230.65.

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